Input function must return 'double' or 'single' values. Found 'sym'.

dear all

when i using integral command although i define z i get the following error

clc
clear all

E2=10;
E1=20;
lE=E2/E1;
n=1;
h=2;
v1=30;
v2=40;
lv=v2/v1;
sym z
Eb=1+(((2*z+h)/2*h)^n)*(lE-1) 
vb=1+(((2*z+h)/2*h)^n)*(lv-1) 

fun=@(z) Eb/((1-(v1^2)*(vb^2))*h)
A11=integral(fun,-1/2,1/2)

and the error :

Undefined function or variable 'z'.

Error in first (line 13)
Eb=1+(((2*z+h)/2*h)^n)*(lE-1)

and also when i use syms z instead of sym z i get the following one :

fun =

  function_handle with value:

    @(z)Eb/((1-(v1^2)*(vb^2))*h)

Error using integralCalc/finalInputChecks (line 511)
Input function must return 'double' or 'single' values. Found 'sym'.

Error in integralCalc/iterateScalarValued (line 315)
                finalInputChecks(x,fx);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in first (line 18)
A11=integral(fun,-1/2,1/2)

i really appreciated if someone could help me with this

Answer 1

[Ameer Hamza]

Correct syntax is to use 'syms', not 'sym'. Also you should use matlabFunction to convert symbolic equation to numeric

clc
clear all
E2=10;
E1=20;
lE=E2/E1;
n=1;
h=2;
v1=30;
v2=40;
lv=v2/v1;
syms z
Eb=1+(((2*z+h)/2*h)^n)*(lE-1) 
vb=1+(((2*z+h)/2*h)^n)*(lv-1) 
fun= matlabFunction(Eb/((1-(v1^2)*(vb^2))*h))
A11=integral(fun,-1/2,1/2)

[shahin hashemi]

dear ameer tanx alot about ur answer

i try it and it work but why when i use equal E1 and E2 (E1=E2) i get the following error :

Error using symengine>@()-3.126954346466542e-4
Too many input arguments.

Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] =
            iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in first (line 24)
A11=integral(fun,-1/2,1/2)

i should mentioned that when i use E1=E2

fun=1

[Ameer Hamza]

shahin, can you paste your code here. I cannot understand how you made E1=E2.

[shahin hashemi]

clc
clear all
E2=20;
E1=20;
lE=E2/E1;
n=1;
h=2;
v1=30;
v2=40;
lv=v2/v1;
syms z
Eb=1+(((2*z+h)/2*h)^n)*(lE-1) 
vb=1+(((2*z+h)/2*h)^n)*(lv-1) 
fun= matlabFunction(Eb/((1-(v1^2)*(vb^2))*h))
A11=integral(fun,-1/2,1/2)

as u say

[Ameer Hamza]

Shahin, running your code, I don't get any error, I get the following output

Eb =
1
 
vb =
(2*z)/3 + 5/3
 
fun =
  function_handle with value:
    @(z)-1.0./((z.*(2.0./3.0)+5.0./3.0).^2.*1.8e+3-2.0)

A11 =
    -2.084250038383697e-04

Can you paste the value of 'fun' after running your code. It should be a function handle, paste the output from command window.

[shahin hashemi]

clc
clear all
d=10;
E2=30;
E1=30;
lE=E2/E1;
n=1;
h=2;
v1=40;
v2=40;
lv=v2/v1;
l1=30;
l2=30;
ll=l2/l1;
i=2;
syms z 
Eb=(1+(((2*z+h)/2*h)^n)*(lE-1)) 
vb=(1+(((2*z+h)/2*h)^n)*(lv-1)) 
lb=(1+(((2*z+h)/2*h)^n)*(ll-1))

fun= matlabFunction(Eb/((1-(v1^2)*(vb^2))*h))
A11=integral(fun,-1/2,1/2)

this is what i exactly try to run

and as u say this is what i exactly get :

fun =

  function_handle with value:

    @()-3.126954346466542e-4

Error using symengine>@()-3.126954346466542e-4
Too many input arguments.

Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] =
            iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in first (line 25)
A11=integral(fun,-1/2,1/2)

im really sorry to bother u and really appreciate about your help

[shahin hashemi]

and i should mention that i can run this cod like this :

fun=@(z) (1+(((2*z+h)/2*h)^n)*(lE-1))/(1-(v1.^2).*((1+(((2*z+h)/2*h)^n)*(lv-1))^2)*h);
A11=integral(fun,-1/2,1/2,'ArrayValued',true)

[Ameer Hamza]

Shahin, for this case, your function is just a constant value and not a function of z, this created issue with the integral function. In this case, try

d=10;
E2=30;
E1=30;
lE=E2/E1;
n=1;
h=2;
v1=40;
v2=40;
lv=v2/v1;
l1=30;
l2=30;
ll=l2/l1;
i=2;
syms z 
Eb=(1+(((2*z+h)/2*h)^n)*(lE-1)) 
vb=(1+(((2*z+h)/2*h)^n)*(lv-1)) 
lb=(1+(((2*z+h)/2*h)^n)*(ll-1))

fun = matlabFunction(Eb/((1-(v1^2)*(vb^2))*h))
A11=integral(@(x) fun(),-1/2,1/2, 'ArrayValued', 1)

[shahin hashemi]

aha got

as i usnderstand we cant have cod that handle both situation

tanx a lot again

[shahin hashemi]

d=10;
E2=30;
E1=30;
lE=E2/E1;
n=1;
h=2;
v1=40;
v2=40;
lv=v2/v1;
l1=30;
l2=30;
ll=l2/l1;
i=2;
syms z 
Eb=(1+(((2*z+h)/2*h)^n)*(lE-1)) 
vb=(1+(((2*z+h)/2*h)^n)*(lv-1)) 
lb=(1+(((2*z+h)/2*h)^n)*(ll-1))
if E1==E2
fun = matlabFunction(Eb/((1-(v1^2)*(vb^2))*h))
A11=integral(@(x) fun(),-1/2,1/2, 'ArrayValued', 1)
else 
    fun= matlabFunction(Eb/((1-(v1^2)*(vb^2))*h))
    A11=integral(fun,-1/2,1/2)
end

i use like this

[Ameer Hamza]

Your solution is correct. You can also use this to avoid if-else block

d=10;
E2=30;
E1=10;
lE=E2/E1;
n=1;
h=2;
v1=40;
v2=40;
lv=v2/v1;
l1=30;
l2=30;
ll=l2/l1;
i=2;
syms z 
Eb=(1+(((2*z+h)/2*h)^n)*(lE-1)) 
vb=(1+(((2*z+h)/2*h)^n)*(lv-1)) 
lb=(1+(((2*z+h)/2*h)^n)*(ll-1))
sum_fun(z) = Eb/((1-(v1^2)*(vb^2))*h); % this lines forces it to act as a function of z, even if the value is constant
fun = matlabFunction(sum_fun)
A11=integral(fun,-1/2,1/2, 'ArrayValued', 1)

[shahin hashemi]

wow you are my hero :D tanx

[Ameer Hamza]

Glad to be of help :D